2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

a. Với m=6 thì phương trình (1) có dạng 

x^2 - 5x +4= 0

<=> (x-1)(x-4)=0

<=> x=1 hoặc x=4

Vậy m=6 thì phương trình có nghiệm x=1 hoặc x=4

b. Xét \(\text{ Δ}=\left(-5\right)^2-4\cdot1\cdot\left(m-2\right)=33-4m\)

Để (1) có nghiệm phân biệt khi \(m< \dfrac{33}{4}\)

Theo Vi-et ta có: \(x_1x_2=m-2;x_1+x_2=5\)

Để 2 nghiệm phương trình (1) dương khi m>2

Ta có:

\(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{5}{m-2}+\dfrac{2}{\sqrt{m-2}}=\dfrac{9}{4}\Leftrightarrow20+8\sqrt{m-2}=9\left(m-2\right)\\ \Leftrightarrow\left(\sqrt{m-2}-2\right)\left(9\sqrt{m-2}+10\right)=0\Leftrightarrow\sqrt{m-2}=2\Leftrightarrow m-2=4\Leftrightarrow m=6\left(t.m\right)\)

ĐKXĐ:

\(\left(2x+2-2\sqrt{5x-1}\right)+\left(\sqrt{5x^2+x+3}-\left(2x+1\right)\right)+x^2-3x+2=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+2\right)}{x+1+\sqrt{5x-1}}+\dfrac{x^2-3x+2}{\sqrt{5x^2+x+3}+2x+1}+x^2-3x+2=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\dfrac{2}{x+1+\sqrt{5x-1}}+\dfrac{1}{\sqrt{5x^2+x+3}+2x+1}+1\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\)

ĐKXĐ: \(x\ge1\)

\(\sqrt{5x-1}=\sqrt{3x-2}+\sqrt{x-1}\)

\(\Leftrightarrow5x-1=3x-2+x-1+2\sqrt{\left(3x-2\right)\left(x-1\right)}\)

\(\Leftrightarrow x+2=2\sqrt{\left(3x-2\right)\left(x-1\right)}\)

\(\Leftrightarrow x^2+4x+4=4\left(3x-2\right)\left(x-1\right)\)

\(\Leftrightarrow11x^2-24x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{11}\left(loại\right)\\x=2\end{matrix}\right.\)